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Difference between revisions of "2002 AMC 12B Problems/Problem 1"
 
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==Solution 3==
 
==Solution 3==
−
The arithmetic mean is <math>\frac{(10^1-1)+(10^2-2)+\ldots+(10^9-1)}{9}=\frac{1111111101}{9}=123456789</math>. So select <math>\boxed{\mathrm{A}}</math>.
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The arithmetic mean is <math>\frac{(10^1-1)+(10^2-1)+\ldots+(10^9-1)}{9}=\frac{1111111101}{9}=123456789</math>. So select <math>\boxed{\mathrm{A}}</math>.
 
~hastapasta
 
~hastapasta
 +
 +
==Video Solution by Daily Dose of Math==
 +
 +
https://youtu.be/w0Q7LG7jdg8
 +
 +
~Thesmartgreekmathdude
 +
 
== See also ==
 
== See also ==
 
{{AMC10 box|year=2002|ab=B|num-b=2|num-a=4}}
 
{{AMC10 box|year=2002|ab=B|num-b=2|num-a=4}}

Latest revision as of 22:21, 24 October 2024

The following problem is from both the 2002 AMC 12B #1 and 2002 AMC 10B #3, so both problems redirect to this page.

Problem

The arithmetic mean of the nine numbers in the set $\{9, 99, 999, 9999, \ldots, 999999999\}$ is a $9$-digit number $M$, all of whose digits are distinct. The number $M$ doesn't contain the digit

$\mathrm{(A)}\ 0 \qquad\mathrm{(B)}\ 2 \qquad\mathrm{(C)}\ 4 \qquad\mathrm{(D)}\ 6 \qquad\mathrm{(E)}\ 8$

Solution 1

We wish to find $\frac{9+99+\cdots +999999999}{9}$, or $\frac{9(1+11+111+\cdots +111111111)}{9}=123456789$. This doesn't have the digit 0, so the answer is $\boxed{\mathrm{(A)}\ 0}$

Solution 2

Notice that the final number is guaranteed to have the digits $\{1, 3, 5, 7, 9\}$ and that each of these digits can be paired with an even number adding up to 9. $\boxed{\mathrm{(A)}\ 0}$ can be taken out, with the other digits fulfilling divisibility by 9.

Solution 3

The arithmetic mean is $\frac{(10^1-1)+(10^2-1)+\ldots+(10^9-1)}{9}=\frac{1111111101}{9}=123456789$. So select $\boxed{\mathrm{A}}$. ~hastapasta

Video Solution by Daily Dose of Math

https://youtu.be/w0Q7LG7jdg8

~Thesmartgreekmathdude

See also

2002 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions
2002 AMC 12B (Problems • Answer Key • Resources)
Preceded by
First question 		Followed by
Problem 2
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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